Answer : $\large\frac{w}{w(1+2 \alpha \bigtriangleup t)}$
Explanation :
Since the angular momentum is conserved ,
$I_{1}w_{1}=I_{2}w_{2}$
$\large\frac{ml^2}{12}w=\large\frac{m(l+\bigtriangleup l)^2}{12} w_{2}$
$\large\frac{ml^2}{12}w=\large\frac{ml^2}{12}(1+\large\frac{\bigtriangleup l}{l})^2 w_{2}$
$w \approx (1+2 \large\frac{2 \bigtriangleup l}{l})w_{2}$
$[\large\frac{\bigtriangleup l}{l} \to 0]$
$w \approx (1 + 2 \alpha \bigtriangleup t) w_{2} $
$w_{2} \approx \large\frac{w}{(1+2 \alpha \bigtriangleup)}$